How to Calculate Work on a Satellite in a Gravitational Field

[Momentum09]

Homework Statement

Given: The universal gravitational constant G = 6.67 E-11, the mass of the Earth M = 5.98E24, and its radius R = 6.7 E6. How much work must an external force do on the satellite to move it from a circular orbit of radius 2R to 3R, if its mass is 2000kg? Answer in Joules.

Homework Equations

I know that E = -GMm/2r, where M is the mass of the Earth and m = satellite.

The Attempt at a Solution

So I calculated the total energy in each radii,
for 2R, -GMm/4R
for 3R, -GMm/6R
then I subtract one from the other to get the net energy change. I don't know what to do after that. Please help. Thank you!

 

[Kurdt]

There is a considerably simpler way of doing this. The work done in moving an object from one orbit to another is:

$$W=\int_{r_1}^{r_2} F(r) dr$$

EDIT: Sorry forgot to say that your method is fine for potential energy not total mechanical energy as you have used.

 

[Momentum09]

thank you so much! :)

 

[Dick]

There is a considerably simpler way of doing this. The work done in moving an object from one orbit to another is:

$$W=\int_{r_1}^{r_2} F(r) dr$$

EDIT: Sorry forgot to say that your method is fine for potential energy not total mechanical energy as you have used.

Actually, I think he is ok. The kinetic energy at radius R is GMm/(2R). The potential is -GMm/R. So the sum is, as he states, -GMm/(2R).

 

[Kurdt]

Was just the procedure in the attempt at the solution that made me think it was total mechanical energy.

 

[ttt359]

I have the exact same question. This is one of the method that I used:
E intial = Ki + Ui = 0 + (-GMm/2R)
E final = Kf + Uf = (1/2)(GMm/3R) + (-GMm/3R)
then Work = E final - E intial

this make sense right? Then why it doesn't work

I also try the integrate method, but nothing work! Can somebody please tell me why?
and how to fix it?

 

[Dick]

If it's in orbit at radius 2R, how can Ki=0?

 

[ttt359]

ok, so Ki is not = O; Then why is it the integration method didn't work either?

 

[Dick]

They both work. You just have to do them right.